Mole concept There are 6,02 x 10 23 atoms in 12 g of carbon-12. - PowerPoint PPT Presentation

Example: Use the unbalanced chemical equation to answer the questions following: 2SO 2 (g) + O 2 (g) → 2SO 3 (g) 3.5 If 160 g of SO 2 reacts with sufficient O 2 , calculate the mass of SO 3 formed. From the balanced equation 2,5 mol of SO 2 will form 2,5 mol of SO 3 M(SO 3 ) = 80 g.mol -1 𝑛 n = 𝑁 𝑛 → m = 200 g SO 3 2,5 = 80

Example: Use the unbalanced chemical equation to answer the questions following: 2SO 2 (g) + O 2 (g) → 2SO 3 (g) 3.6 How many molecules of SO 3 are formed in 3.5? 1 mol of molecules is 6,02 x 10 23 molecules 2,5 mol of molecules is n x N A = 2,5 x 6,02 x 10 23 molecules = 1,505 x 10 24 SO 3 molecules

Example: Use the unbalanced chemical equation to answer the questions following: 2SO 2 (g) + O 2 (g) → 2SO 3 (g) 3.7 Is it possible to produce 80 g of SO 3 if only 8 g of O 2 is available? Explain. M(SO 3 ) = 80 g.mol -1 80 𝑛 𝑁 = 80 = 1 mol n =

Example: Use the unbalanced chemical equation to answer the questions following: 2SO 2 (g) + O 2 (g) → 2SO 3 (g) 3.7 Is it possible to produce 80 g of SO 3 if only 8 g of O 2 is available? Explain. From the balanced equation 1 mol of O 2 is needed to form 2 mol of SO 3 → 0,5 mol of O 2 is needed to form 1 mol of SO 3 M(O 2 ) = 32 g.mol -1 𝑛 n = 𝑁 𝑛 → m = 16 g 0,5 = 32 → 16 g of oxygen is needed, therefore 8 g will not be sufficient.

In 3.7 oxygen will be the limited reagent and SO 2 will be in excess if there is more than 1 mol of SO 2 available.

Percentage yield 𝑏𝑑𝑢𝑣𝑏𝑚 𝑧𝑗𝑓𝑚𝑒 Percentage yield = 𝑢ℎ𝑓𝑝𝑠𝑓𝑢𝑗𝑑𝑏𝑚 𝑧𝑗𝑓𝑚𝑒

Example, pg. 167 When 12,8 g of Cu is allowed to burn (combust) in oxygen, 15,2 g of copper(II) oxide is produced. Determine the percentage yield. 2Cu(s) + O 2 (g) → 2CuO(s) M(CuO) = 79,5 g.mol -1 M(Cu) = 63,5 g.mol -1 𝑛 12,8 Cu: n = 𝑁 = 63,5 = 0,2 mol Cu From the balanced equation: 2 mol of Cu yields 2 mol of CuO 0,2 mol of Cu yields 0,2 mol of CuO

Example, pg. 167 When 12,8 g of Cu is allowed to burn (combust) in oxygen, 15,2 g of copper(II) oxide is produced. Determine the percentage yield. 2Cu(s) + O 2 (g) → 2CuO(s) 𝑛 n = 𝑁 𝑛 0,2 = 79,5 m = 16,03 g (theoretical yield) 𝑏𝑑𝑢𝑣𝑏𝑚 𝑧𝑗𝑓𝑚𝑒 Percentage yield = 𝑢ℎ𝑓𝑝𝑠𝑓𝑢𝑗𝑑𝑏𝑚 𝑧𝑗𝑓𝑚𝑒 x 100 15,2 16,03 x 100 = = 94,82 %

Applications of exothermic reactions producing large quantities of gas: Airbags: 2NaN 3 (s) → 2Na(s) + 3N 2 (g) Only 130 g of NaN 3 is needed to produce 67 dm 3 of N 2 gas in 0,03 s.

Applications of exothermic reactions producing large quantities of gas: Octane combustion in motorcar engine: 2C 8 H 18 + 25O 2 → 16CO 2 + 18H 2 O + energy The gas moves the pistons in the engine.

Applications of exothermic reactions producing large quantities of gas: Termionic decomposition of ammonium nitrate: 2NH 4 NO 3(s) → 2N 2(g) + 4H 2 O(g) + O 2 (g) Fertiliser production and explosive in mining industry.

Homework : Work through the following questions and answers. Always FIRST try to do the question YOURSELF.

1. Balance the following equation: K + O 2 → 4 2 K 2 O Now use the balanced equation to complete the following table with each row in the table according to the same ratio as the balanced equation: 4K O 2 2K 2 O 0,5 2 1 1,6 0,4 0,8 0,3 0,6 0,15 0,25 1 0,5 10 2,5 5

2. Write down the balanced equation for nitrogen reacting with hydrogen to form ammonia: N 2 + H 2 → 2 3 NH 3 Now use the balanced equation to complete the following table with each row in the table according to the same ratio as the balanced equation: N 2 3H 2 2NH 3 1,5 4,5 3 0,4 0,2 0,6 2 1 3 0,9 0,3 0,6 10 30 20

3. Consider the following reaction: NH 3 + O 2 → NO + H 2 O 3.1 Balance the reaction equation. NH 3 + O 2 → 4 4 5 NO + H 2 O 6

3. Consider the following reaction: 4NH 3 + 5O 2 → 4NO + 6H 2 O 3.2 How many moles of oxygen is needed to react with 1 mole of ammonia? 4 mol ammonia reacts with 5 mol oxygen 1 4 × 5 ) = 1,25 mol oxygen 1 mol ammonia reacts with (

3. Consider the following reaction: 4NH 3 + 5O 2 → 4NO + 6H 2 O 3.3 What mass of oxygen reacted with 1 mole of ammonia according to question 3.2? 𝑛 n = M(O 2 ) = 32 g.mol -1 𝑁 𝑛 1,25 = 32 m = 40 g

4. Ammonia is commercially prepared as follows in the Haber process: N 2 + 3H 2 → 2NH 3 4.1 How many moles of NH 3 is formed if one mole of N 2 reacts with sufficient hydrogen? 2 mol NH 3

4. Ammonia is commercially prepared as follows in the Haber process: N 2 + 3H 2 → 2NH 3 4.2 How many moles of ammonia is formed when 1,806 x 10 24 molecules of hydrogen reacts with sufficient nitrogen? 𝑜𝑣𝑛𝑐𝑓𝑠 𝑝𝑔 𝑞𝑏𝑠𝑢𝑗𝑑𝑚𝑓𝑡 n = 𝑂 𝐵 1,806×10 24 = 6,02×10 23 = 3 mol H 2 3 mol of H 2 reacts with nitrogen to form 2 mol of NH 3

4. Ammonia is commercially prepared as follows in the Haber process: N 2 + 3H 2 → 2NH 3 4.3 How many moles of ammonia is formed if 0,2 mol of N 2 combines with 0,6 mol of H 2 ? 1 mol of N 2 reacts with 3 mol H 2 to form 2 mol of NH 3 0,2 mol of N 2 reacts with 0,6 mol H 2 to form 0,4 mol of NH 3

N 2 + 3H 2 → 2NH 3 4.4 How many moles of ammoniais formed from 6 moles of H 2 ? 3 mol of H 2 reacts to form 2 mol of NH 3 6 mol of H 2 reacts to form 4 mol of NH 3

N 2 + 3H 2 → 2NH 3 4.5 Calculate the volume of gas which would form from 6 mol of H 2 at STP? 4 mol of NH 3 forms 𝑊 n = 𝑊 𝑛 𝑊 4 = 22,4 V = 89,6 dm 3

N 2 + 3H 2 → 2NH 3 4.6 What mass of nitrogen is needed to form 8,5 g of ammonia? M(NH 3 ) = 17 g.mol -1 𝑛 8,5 n = 𝑁 = 17 = 0,5 mol 2 mol of ammonia is formed from 1 mol of nitrogen 0,5 mol of ammonia is formed from 0,25 mol of nitrogen 𝑛 n = M(N 2 ) = 28 g.mol -1 𝑁 𝑛 28 → m = 7 g 0,25 =

N 2 + 3H 2 → 2NH 3 4.7 What is the volume of 8,5 g of ammonia at STP? 𝑊 n = 𝑊 𝑛 𝑊 0,5 = 22,4 V = 11,2 dm 3

5. Methanol (CH 3 OH) can be burnt in oxygen to provide energy or it can decompose into hydrogen gas which can be used as fuel, according to the following reaction: CH 3 OH(ℓ) → 2H 2 (g) + CO(g) 5.1 If 125 g of methanol decomposes, what is the theoretical yield of the hydrogen gas? n methanol : n H ₂ M(CH 3 OH) = 32 g.mol -1 1 : 2 𝑛 125 n = 𝑁 = = 3,91 mol 3,91 : 7,81 32 𝑛 n H ₂ = 𝑁 𝑛 7,81 = 2 m = 15,63 g

5. Methanol (CH 3 OH) can be burnt in oxygen to provide energy or it can decompose into hydrogen gas which can be used as fuel, according to the following reaction: CH 3 OH(ℓ) → 2H 2 (g) + CO(g) 5.2 If only 13,6 g of hydrogen gas is obtained, what was the percentage yield? 13,6 % yield = 15,63 x 100 = 87 %

6. Consider the following balanced equation: 2Aℓ(s ) + 3 Cℓ 2 (g) → 2 AℓCℓ 3 (s) A mixture of 1,5 moles of Aℓ and 3 moles of Cℓ 2 are allowed to react together. 6.1 Which substance is the limiting reactant? 2 mol of Aℓ reacts with 3 mol of chlorine gas 1,5 mol of Aℓ reacts with 2,25 mol of chlorine gas OR Aℓ : Cℓ 2 2 : 3 1,5 : 2,25 3 mol of chlorine gas is available and it is therefore in excess. Therefore the aluminium is the limiting reactant.

6. Consider the following balanced equation: 2Aℓ(s) + 3 Cℓ 2 (g) → 2 AℓCℓ 3 (s) A mixture of 1,5 moles of Aℓ and 3 moles of Cℓ 2 are allowed to react together. 6.2 How many moles of AℓCℓ 3 is formed? 2 mol of Aℓ forms 2 mol of AℓCℓ 3 1,5 mol of Aℓ forms 1,5 mol of AℓCℓ 3 OR Aℓ : Cℓ 2 : AℓCℓ 3 2 : 3 : 2 1,5 : 2,25 : 1,5 thus 1,5 mol of AℓCℓ 3 can be produced

6. Consider the following balanced equation: 2Aℓ(s) + 3 Cℓ 2 (g) → 2 AℓCℓ 3 (s) A mixture of 1,5 moles of Aℓ and 3 moles of Cℓ 2 are allowed to react together. 6.3 How many mole of the reactant that was in excess, was eventually left over? There is 3 mol but only 2,25 mol of Cℓ 2 is used, thus 3 mol – 2,25 mol = 0,75 mol Cℓ 2

7. Gerda heats 50 g of CaCO 3 which reacts as follows: CaCO 3 → CaO + CO 2 7.1 Determine the mass of gas she captured if the CaCO 3 was 100% pure. M(CH 3 OH) = 100 g.mol -1 𝑛 50 n = 𝑁 = 100 = 0,5 mol 1 mol of CaCO 3 forms 1 mol of CO 2 𝑛 n = 0,5 mol of CaCO 3 forms 0,5 mol of CO 2 𝑁 𝑛 0,5 = 44 m = 22 g

7. Gerda heats 50 g of CaCO 3 which reacts as follows: CaCO 3 → CaO + CO 2 7.2 Gerda only obtains 18,2 g of CO 2 gas. Calculate the percentage purity of the CaCO 3 . Remember that purity will determine the yield. 𝑏𝑑𝑢𝑣𝑏𝑚 𝑧𝑗𝑓𝑚𝑒 % yield = 𝑢ℎ𝑓𝑝𝑠𝑓𝑢𝑗𝑑𝑏𝑚 𝑧𝑗𝑓𝑚𝑒 18,2 = 22 x 100 = 82,7 %

7. Gerda heats 50 g of CaCO 3 which reacts as follows: CaCO 3 → CaO + CO 2 7.3 Calculate the mass of gas Gerda can capture if the CaCO 3 was only 70% pure. Using the answer in 7.1: mass of CO 2 = 70% x 22 g = 15,4 g OR (if 7.1 was not asked)

7. Gerda heats 50 g of CaCO 3 which reacts as follows: CaCO 3 → CaO + CO 2 7.3 Calculate the mass of gas Gerda can capture if the CaCO 3 was only 70% pure. 70% of 50 g is 35 g 35 𝑛 𝑁 = 100 = 0,35 mol n = 1 mol of CaCO 3 forms 1 mol of CO 2 0,35 mol of CaCO 3 forms 0,35 mol of CO 2 𝑛 n = 𝑁 𝑛 ➔ m = 15,4 g 0,35 = 44

A learner adds 10 cm 3 of sulphuric acid with a concentration of 10 mol.dm -3 to water to 8. make a solution of 250 cm 3 . Calculate the concentration of the diluted acid. 𝑜 c₁ = 𝑊₁ n = 0,01 x 10 n = 0,1 mol sulphuric acid 𝑜 c₂ = 𝑊₂ 0,1 c₂ = 0,25 = 0,4 mol.dm -3

A learner adds 10 cm 3 of sulphuric acid with a concentration of 10 mol.dm -3 to water 8. to make a solution of 250 cm 3 . Calculate the concentration of the diluted acid. Alternative method: 𝑜 𝑗 = 𝑜 𝑔 c 1 V 1 = c 2 V 2 10 x 0,01 = c 2 x 0,25 c 2 = 0,4 mol.dm -3

5. Consider the reaction: Aℓ 2 O 3 + 3H 2 → 2 Aℓ + 3H 2 O Determine the mass of Aℓ 2 O 3 that has to react with an excess of hydrogen to prepare 135 g of Aℓ. M(Aℓ) = 27 g.mol -1 135 𝑛 𝑁 = 27 = 5 mol n = 2 mol of aluminium is formed from 1 mol of Aℓ 2 O 3 5 mol of aluminium is formed from 2,5 mol of Aℓ 2 O 3 M(Aℓ 2 O 3 ) = 102 g.mol -1 𝑛 n = 𝑁 𝑛 2,5 = → m = 255 g 102

9.1 Aℓ : Cℓ 2 2 : 3 1,5 : 2,25 If 1,5 mol of Aℓ is used, only 2,25 mol of Cℓ 2 would be used. Aℓ is the limited reactant. 9.2 Aℓ : Cℓ 2 : AℓCℓ 3 2 : 3 : 2 1,5 : 2,25 : 1,5 thus 1,5 mol of AℓCℓ 3 can be produced 9.3 We have 3 mol, but only 2,25 mol of Cℓ 2 is used, thus 3 – 2,25 = 0,75 mol is left